Post on 28-May-2020
s im i l ar i ty & proport i onal i ty
MPM2D: Principles of Mathematics
Ratios and Proportions
J. Garvin
Slide 1/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A ratio is a comparison between two quantities.
For example, the instructions on a can of orange juiceconcentrate may read “add 1 can of concentrate to 3 cans ofwater.”
The ratio of concentrate to water can be expressed using acolon, 1 : 3, or as a fraction, 1
3 .
Note that this fraction does not mean that the orange juiceis one-third concentrate.
The orange juice is actually one-fourth concentrate, sincethere were four cans added: 1 of concentrate, and 3 of water.
While this often makes it more convenient to express ratiosusing a colon for common measurements, it is usually moreadvantageous for us to use fractions for our calculations.
J. Garvin — Ratios and Proportions
Slide 2/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A proportion is a statement that equates two or more ratios.
For example, 29 = 6
27 is a proportion, since the ratio on theright can be reduced to the ratio on the left.
In general, two ratios are equal if there is some scale factor k ,such that k
k · ab = c
d .
In this case, since 33 · 2
9 = 627 , k = 3.
Many proportions can be solved by identifying the scalefactor, then working out any missing values.
J. Garvin — Ratios and Proportions
Slide 3/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A proportion is a statement that equates two or more ratios.
For example, 29 = 6
27 is a proportion, since the ratio on theright can be reduced to the ratio on the left.
In general, two ratios are equal if there is some scale factor k ,such that k
k · ab = c
d .
In this case, since 33 · 2
9 = 627 , k = 3.
Many proportions can be solved by identifying the scalefactor, then working out any missing values.
J. Garvin — Ratios and Proportions
Slide 3/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A proportion is a statement that equates two or more ratios.
For example, 29 = 6
27 is a proportion, since the ratio on theright can be reduced to the ratio on the left.
In general, two ratios are equal if there is some scale factor k ,such that k
k · ab = c
d .
In this case, since 33 · 2
9 = 627 , k = 3.
Many proportions can be solved by identifying the scalefactor, then working out any missing values.
J. Garvin — Ratios and Proportions
Slide 3/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A proportion is a statement that equates two or more ratios.
For example, 29 = 6
27 is a proportion, since the ratio on theright can be reduced to the ratio on the left.
In general, two ratios are equal if there is some scale factor k ,such that k
k · ab = c
d .
In this case, since 33 · 2
9 = 627 , k = 3.
Many proportions can be solved by identifying the scalefactor, then working out any missing values.
J. Garvin — Ratios and Proportions
Slide 3/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A proportion is a statement that equates two or more ratios.
For example, 29 = 6
27 is a proportion, since the ratio on theright can be reduced to the ratio on the left.
In general, two ratios are equal if there is some scale factor k ,such that k
k · ab = c
d .
In this case, since 33 · 2
9 = 627 , k = 3.
Many proportions can be solved by identifying the scalefactor, then working out any missing values.
J. Garvin — Ratios and Proportions
Slide 3/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
15=
2
5.
Since 5 × 3 = 15, k = 3.
Therefore, x = 2 × k = 2 × 3 = 6.
Example
Solvex
12=
5
8.
Since 8 × 32 = 12, k = 3
2 .
Therefore, x = 5 × k = 5 × 32 = 15
2 .
J. Garvin — Ratios and Proportions
Slide 4/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
d
a
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
A useful technique, known as cross-multiplication, can beused to eliminate two different denominators simultaneously.
a
b=
c
da
b· b =
c
d· b
a =bc
d
a · d =bc
d· d
a · d = b · c
Cross-Multiplication
Ifa
b=
c
d, then a · d = b · c .
J. Garvin — Ratios and Proportions
Slide 5/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
88x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
8
8x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
88x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
88x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
88x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
12=
5
8using cross-multiplication.
x
12=
5
88x = 12 · 5
8x = 60
x = 608
x = 152
J. Garvin — Ratios and Proportions
Slide 6/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
4=
5
7.
x
4=
5
77x = 4 · 5
7x = 20
x = 207
J. Garvin — Ratios and Proportions
Slide 7/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
4=
5
7.
x
4=
5
7
7x = 4 · 5
7x = 20
x = 207
J. Garvin — Ratios and Proportions
Slide 7/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
4=
5
7.
x
4=
5
77x = 4 · 5
7x = 20
x = 207
J. Garvin — Ratios and Proportions
Slide 7/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
4=
5
7.
x
4=
5
77x = 4 · 5
7x = 20
x = 207
J. Garvin — Ratios and Proportions
Slide 7/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex
4=
5
7.
x
4=
5
77x = 4 · 5
7x = 20
x = 207
J. Garvin — Ratios and Proportions
Slide 7/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
3
3(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Solving Proportions
Example
Solvex + 1
8=
4
3.
Be sure to use the distributive law when cross-multiplying.
x + 1
8=
4
33(x + 1) = 8 · 4
3x + 3 = 32
3x = 29
x = 293
J. Garvin — Ratios and Proportions
Slide 8/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Many types of word problems involve proportions.
Information will usually be given about one ratio, while theremaining ratio will require solving.
Proportion problems usually have the format “if a is to b,then c is to. . .”
There may be more than one way to solve the problem. Aslong as the ratios are valid, solving a proportion using theprevious techniques should work.
J. Garvin — Ratios and Proportions
Slide 9/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Many types of word problems involve proportions.
Information will usually be given about one ratio, while theremaining ratio will require solving.
Proportion problems usually have the format “if a is to b,then c is to. . .”
There may be more than one way to solve the problem. Aslong as the ratios are valid, solving a proportion using theprevious techniques should work.
J. Garvin — Ratios and Proportions
Slide 9/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Many types of word problems involve proportions.
Information will usually be given about one ratio, while theremaining ratio will require solving.
Proportion problems usually have the format “if a is to b,then c is to. . .”
There may be more than one way to solve the problem. Aslong as the ratios are valid, solving a proportion using theprevious techniques should work.
J. Garvin — Ratios and Proportions
Slide 9/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Many types of word problems involve proportions.
Information will usually be given about one ratio, while theremaining ratio will require solving.
Proportion problems usually have the format “if a is to b,then c is to. . .”
There may be more than one way to solve the problem. Aslong as the ratios are valid, solving a proportion using theprevious techniques should work.
J. Garvin — Ratios and Proportions
Slide 9/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
5
5s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
If it costs $48.00 to buy 5 shirts, how much should it cost tobuy a dozen shirts?
Set up a proportion comparing the number of shirts, s, totheir costs.
s
12=
48
55s = 12 · 48
5s = 576
s = 115.2
A dozen shirts should cost $115.20.
J. Garvin — Ratios and Proportions
Slide 10/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 800
5 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.
J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Word Problems Involving Proportions
Example
A pallet of bricks has a mass of 10 000 kg. After 1 680 brickshave been used, the pallet has a mass of 5 800 kg. Howmany bricks were on the pallet?
Set up a proportion comparing the number of bricks, b, totheir masses.
b
10 000=
b − 1 680
5 8005 800b = 10 000(b − 1 680)
5 800b = 10 000b − 16 800 000
4 200b = 16 800 000
b = 4 000
A total of 4 000 bricks were on the pallet.J. Garvin — Ratios and Proportions
Slide 11/12
s im i l ar i ty & proport i onal i ty
Questions?
J. Garvin — Ratios and Proportions
Slide 12/12